One mole of an ideal gas undergoes two different cyclic processes I and II, as shown in the 𝑃-𝑉 diagrams below. In cycle I, processes π‘Ž, 𝑏, 𝑐 and 𝑑 are isobaric, isothermal, isobaric and isochoric, respectively. In cycle II, processes π‘Ž ′ , 𝑏 ′ , 𝑐 ′ and 𝑑 ′ are isothermal, isochoric, isobaric and isochoric, respectively. The total work done during cycle I is π‘ŠπΌ and that during cycle II is π‘ŠπΌπΌ. The ratio π‘ŠπΌ/π‘ŠπΌπΌ is _______.

 To determine the ratio 

WIWII of the work done during cycle I to that during cycle II, we need to analyze each cycle step-by-step and calculate the work done in each process.

Cycle I:

  1. Process a (Isobaric Expansion):

    • Work done: Wa=P1(V2V1)

  2. Process b (Isothermal Compression):

    • Work done: Wb=nRT2ln(V1V2)

  3. Process c (Isobaric Compression):

    • Work done: Wc=P2(V1V2)

  4. Process d (Isochoric Process):

    • Work done: Wd=0 (since volume is constant)

Total Work for Cycle I:

WI=Wa+Wb+Wc+WdWI=P1(V2V1)+nRT2ln(V1V2)+P2(V1V2)

Cycle II:

  1. Process a (Isothermal Expansion):

    • Work done: Wa=nRT1ln(V2V1)

  2. Process b (Isochoric Process):

    • Work done: Wb=0 (since volume is constant)

  3. Process c (Isobaric Compression):

    • Work done: Wc=P2(V1V2)

  4. Process d (Isochoric Process):

    • Work done: Wd=0 (since volume is constant)

Total Work for Cycle II:

WII=Wa+Wb+Wc+WdWII=nRT1ln(V2V1)+P2(V1V2)

Ratio WIWII:

WIWII=P1(V2V1)+nRT2ln(V1V2)+P2(V1V2)nRT1ln(V2V1)+P2(V1V2)

This ratio depends on the specific values of P1,P2,V1,V2,T1, and T2. Without numerical values, we cannot simplify this expression further. However, the ratio WIWII can be calculated if the specific values of these parameters are provided.

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